What is the slope of the line tangent to $f(x) = x^{2}-x-8$ at $x = -3$ ?
Explanation: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{((x+\Delta x)^{2}-(x+\Delta x)-8) - (x^{2}-x-8)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(x^{2}+2x \Delta x+\Delta x^{2}-(x+\Delta x)-8) - (x^{2}-x-8)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{x^{2}+2(x \Delta x)+\Delta x^{2}-x-\Delta x-8-x^{2}+x+8}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{2(x \Delta x)+\Delta x^{2}-\Delta x}{\Delta x}$ $ = \lim_{\Delta x \to 0} 2x+\Delta x-1$ $ = 2x-1$ $ = (2)(-3)-1$ $ = -7$